Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
public class Main
{

	public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		double a[]=new double[n+1];
		for(int i=1;i<=n;i++)
		{
			a[i]=sc.nextDouble();
		}
		double dp[][]=new double[n+1][n+1];
		dp[0][0]=1;
		for(int i=1;i<=n;i++)
		{
			for(int j=0;j<=i;j++)
			{
				if(j==0)
					dp[i][j]= (dp[i-1][j]*(1-a[i]));
				else
				dp[i][j]=(dp[i-1][j-1]*a[i]) + (dp[i-1][j]*(1-a[i]));
			}
		}
		double sum=0;
		for(int i=(n+1)/2;i<=n;i++)
			sum+=dp[n][i];
		System.out.println(sum);
	}
}